Theo đề bài ta có : nHCl = 0,2.0,2=0,04(mol)

a) Ta có PTHH :

\(HCl+NaOH\rightarrow NaCl+H2O\)

0,04mol.....0,04mol....0,04mol

Ta có :

\(V_{\text{dd}HCl\left(c\text{ần}-d\text{ùng}\right)}=\dfrac{0,04}{0,1}=0,4\left(lit\right)=400\left(ml\right)\)

CM_NaCl = \(\dfrac{0,04}{0,2}=0,2\left(M\right)\)

b) Theo đề bài ta có : mddHCl=\(200.1=200\left(g\right)\)

Ta có PTHH :

\(Ca\left(OH\right)2+2HCl\rightarrow CaCl2+2H2O\)

0,02mol...........0,04mol....0,02mol

Ta có :

\(m\text{dd}Ca\left(OH\right)2\left(c\text{ần}-d\text{ùng}\right)=\dfrac{0,02.74}{5}.100=29,6\left(g\right)\)

C%_CaCl2 = \(\dfrac{0,02.111}{0,02.74+200}.100\%\approx1,102\%\)

Vậy..............

Làm nhanh zùm mk

a)PTHH:
HCl + NaOH → NaCl + H2O
nHCl = 0,04 (mol) = nNaOH = nNaCl
=>VddNaOH = 0,04/0,1 = 0,4 (l) = 400 (ml)
Vdd = VddNaOH + VddHCl = 0,6 (l)
=>C(M) ≈ 0,067 (M)
b) 2HCl + Ca(OH)2 → CaCl2 + 2H2O
nCa(OH)2 = nCaCl2 = (1/2)nHCl = 0,02 (mol)
(Nồng độ phần trăm = 25% ????)
mCa(OH)2 = 1,48 (g)
=>mdd(Ca(OH)2) = 5,92 (g)
mddHCl = 220 (g)
=>mdd = 225,92 (g)
mCaCl2 = 2,22 (g)
=>%mCaCl2 ≈ 0,98%

nồng độ phần trăm = 5%

conan cung pai hoi

Bài 1: \(Ca\left(OH\right)_2\left(0,3\right)+2HCl\left(0,6\right)\rightarrow2CaCl_2\left(0,6\right)+2H_2O\)

\(n_{Ca\left(OH\right)_2}=0,3\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,6}{0,2}=3M\)

\(m_{CaCl_2}=0,6.111=66,6\left(g\right)\)

\(C_{MddCaCl_2}=\dfrac{0,6}{0,3}=2M.\)

Bài 2: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(n_{Fe}=0,1\left(mol\right)\)

\(n_{H_2SO_4}=0,1\left(mol\right)\)

=> Pư này pư vừa đủ

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(C_{MddFeSO_4}=\dfrac{0,1}{0,5}=0,2M\)

\(n_{FeSO_4}=n_{FeSO_4.7H_2O}=0,1\left(mol\right)\)

\(\Rightarrow m_{FeSO_4.7H_2O}=0,1.278=27,8\left(g\right)\)

\(4H_2\left(0,1\right)+Fe_3O_4\left(0,025\right)\rightarrow3Fe+4H_2O\)

\(\Rightarrow m_{Fe_3O_4}=0,025.232=5,8\left(g\right).\)

Bài 1: \(n_{H_2SO_4}=\frac{9}{49}\left(mol\right)\)

H₂SO₄ + 2KOH -> K₂SO₄ + 2H₂O

=> n_KOH= 2n_H2SO4 = \(\frac{18}{49}\left(mol\right)\)

=> V_{dd KOH} = \(\frac{18}{49}:\frac{2}{1000}=\frac{9000}{49}\left(ml\right)\)

b) n_K2SO4 = n_H2SO4 = \(\frac{9}{49}\left(mol\right)\)

=> m_K2SO4= \(\frac{9}{49}\cdot174=\frac{1566}{49}\left(g\right)\)

m_{dd KOH} = \(\frac{9000}{49}\cdot1,12=\frac{1440}{7}\left(g\right)\)

c) \(\%m_{K_2SO_4}=\frac{1566}{49}:\left(200+\frac{1440}{7}\right)\cdot100\%\approx7,87\%\)

bài 2: n_Na2CO3 = 0,05 (mol)

PTHH:

Na₂CO₃ + 2HCl -> 2NaCl + H₂O + CO₂

=> n_HCl = n _NaCl = 2n_Na2CO3 = 0,1 (mol)

=> m_NaCl= 0,1 . 58,5 = 5,85 (g)

b) n_CO2 = n_Na2CO3 = 0,05 (mol)

=> m_CO2 = 0,05 . 44 = 2,2 (g)

m_{dd HCl} = 0,1 . 36,5 :20% = 18,25 (g)

=> %m_NaCl = \(\frac{5,85}{53+18,25-2,2}\approx8,47\%\)

nZn = \(\frac{6,5}{65}=0,1\) mol

a/ PTHH : Zn + 2HCl -----> ZnCl2 + H2

                   1                                           1

                  0,1                                       0,1

=> V_H2 = 0,1 x 22,4 = 2,24 (l)

c/ nHCl = 2nZn = 0,2 mol

\(\Rightarrow C_{M_{HCl}}=\frac{0,2}{\frac{100}{1000}}=2M\)

d/\(n_{HCl}\) trong 500ml dd HCl là : \(2\times0,5=1\left(mol\right)\)

PTHH : 2HCl + Ca(OH)₂ -----> CaCl₂ + 2H₂O

                  1             1

Suy ra \(V_{Ca\left(OH\right)_2}=\frac{n_{Ca\left(OH\right)_2}}{C_{M_{Ca\left(OH\right)_2}}}=\frac{1}{2}=0,5\left(l\right)\)

nếu như số mol của HCl=1 mol .... Theo PTHH thì Ca(OH)2 phải là 0,5 mol chứ...????

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

Theo PT:  \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)

a, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)

b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)

c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)

Cảm ơn nha